There are 7 seniors on the student council. Two of them will be chosen to go to an all-district meeting. How many ways are there to choose the students who will go to the meeting?

Decide if this is a permutation or a combination, and then find the number of ways to choose the students who go.

A. Permutation; number of ways [tex]$=42$[/tex]

B. Combination; number of ways [tex]$=42$[/tex]

C. Combination; number of ways [tex]$=21$[/tex]

D. Permutation; number of ways [tex]$=21$[/tex]



Answer :

To determine the number of ways to choose 2 seniors from a group of 7 to go to an all-district meeting, we need to decide if this is a permutation or a combination.

### Step 1: Understand the difference between permutations and combinations

- Permutations: Used when the order of selection matters. For example, if selecting different positions such as President and Vice-President, the order in which the individuals are selected is important.

- Combinations: Used when the order does not matter. If just selecting a group, the order of selection is irrelevant.

Here, we are simply selecting 2 students to attend a meeting, and the order in which they are chosen doesn't matter. Therefore, this is a problem involving combinations.

### Step 2: Select the appropriate formula

For combinations, the formula for choosing [tex]\( r \)[/tex] items from [tex]\( n \)[/tex] items is:

[tex]\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \][/tex]

Where:
- [tex]\( n \)[/tex] is the total number of items.
- [tex]\( r \)[/tex] is the number of items to choose.
- [tex]\( ! \)[/tex] denotes factorial, which means the product of all positive integers up to that number.

### Step 3: Substitute the given values into the formula

- Total number of seniors, [tex]\( n = 7 \)[/tex]
- Number of seniors to be chosen, [tex]\( r = 2 \)[/tex]

So, we substitute [tex]\( n = 7 \)[/tex] and [tex]\( r = 2 \)[/tex]:

[tex]\[ \binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7!}{2! \cdot 5!} \][/tex]

### Step 4: Simplify the factorials

[tex]\[ 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \][/tex]

However, since [tex]\( 5! \)[/tex] is in both the numerator and the denominator, they cancel out:

[tex]\[ \frac{7 \times 6 \times 5!}{2! \times 5!} = \frac{7 \times 6}{2!} \][/tex]

And [tex]\( 2! \)[/tex] is:

[tex]\[ 2! = 2 \times 1 = 2 \][/tex]

So the calculation becomes:

[tex]\[ \frac{7 \times 6}{2 \times 1} = \frac{42}{2} = 21 \][/tex]

### Final Answer

Therefore, the number of ways to choose 2 seniors from 7 is given by the combination, which is [tex]\(\boxed{21}\)[/tex].

The correct answer is: C. Combination; number of ways [tex]\( = 21\)[/tex].